Constructing Roads Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19884   Accepted: 8315 Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each…
题意:给出n个村庄之间的距离,再给出已经连通起来了的村庄.求把所有的村庄都连通要修路的长度的最小值. 思路:Kruskal算法 课本代码: //Kruskal算法 #include<iostream> using namespace std; int fa[120]; int get_father(int x){ return fa[x]=fa[x]==x?x:get_father(fa[x]);//判断两个节点是否属于一颗子树(并查集) } int main(){ int n; int p[…
Constructing Roads Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2421 Appoint description:  System Crawler  (2015-05-27) Description There are N villages, which are numbered from 1 to N, and y…
题目链接:http://poj.org/problem?id=2421 实际上又是考最小生成树的内容,也是用到kruskal算法.但稍稍有点不同的是,给出一些已连接的边,要在这些边存在的情况下,拓展出最小生成树来. 一般来说,过到这四组数据大体上就能AC了.  1.题目给出的案例数据    2.连接的道路可能把所有的村庄都已经连通了       3.两个村庄给出多次,即连接这两个村庄的道路是重复的!. 2.3这两种情况的数据如下(为了好看,自己出了一组,当然Sample Input 那组也行):…
Constructing Roads 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/D Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two villa…
Constructing Roads Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road betw…
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a…
题意:要在n个城市之间建造公路,使城市之间能互相联通,告诉每个城市之间建公路的费用,和已经建好的公路,求最小费用. 解法:最小生成树.先把已经建好的边加进去再跑kruskal或者prim什么的. 代码: #include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<math.h> #includ…
给一个n个点的完全图 再给你m条道路已经修好 问你还需要修多长的路才能让所有村子互通 将给的m个点的路重新加权值为零的边到边集里 然后求最小生成树 #include<cstdio> #include<iostream> #include<algorithm> #include<cmath> #include<cstring> #include<string> #define cl(a,b) memset(a,b,sizeof(a))…
题意:有几个村庄,要修最短的路,使得这几个村庄连通.但是现在已经有了几条路,求在已有路径上还要修至少多长的路. 分析:用Prim求最小生成树,将已有路径的长度置为0,由于0是最小的长度,所以一定会被Prim选中加入最小生成树. package Map; import java.util.Scanner; /** * Prime */ public class Poj_2421_Prim { static int MAXVEX = 200; static int n, m; static int[…
Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or the…
There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a…
Constructing Roads There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B,…
求最小生成树.有一点点的变化,就是有的边已经给出来了.所以,最小生成树里面必须有这些边,kruskal和prim算法都能够,prim更简单一些.有一点须要注意,用克鲁斯卡尔算法的时候须要将已经存在的边预处理一下,并查集转化为同一个祖先.记得要找他们的祖先再转化.普里姆算法仅仅须要将那些已经存在的边都初始化为0就能够了. kruskal: #include<iostream> #include<cstdlib> #include<cstring> #include<…
任意门:http://poj.org/problem?id=1984 Navigation Nightmare Time Limit: 2000MS   Memory Limit: 30000K Total Submissions: 7783   Accepted: 2801 Case Time Limit: 1000MS Description Farmer John's pastoral neighborhood has N farms (2 <= N <= 40,000), usuall…
链接:http://acm.hdu.edu.cn/showproblem.php?pid=4641 题意:有一个长度为n(n < 5e4)的字符串,Q(Q<=2e5)次操作:操作分为:在末尾插入一个字符ch和查询不同子串出现次数不小于K的数量: 思路1:SAM在线求解: 对于每次找到将一个字符x插入到SAM之后,我们知道pre[p]所含有的Tx的后缀字符串数目为step[pre[np]]个,那么只需要每次插入之后更新下这些字符串出现的次数cnt即可: 由于Right(fa)与Right(r)没…
Navigation Nightmare Description Farmer John's pastoral neighborhood has N farms (2 <= N <= 40,000), usually numbered/labeled 1..N. A series of M (1 <= M < 40,000) vertical and horizontal roads each of varying lengths (1 <= length <= 100…
转载:http://www.cnblogs.com/hxer/p/5675149.html 题意:有一个长度为n(n < 5e4)的字符串,Q(Q<=2e5)次操作:操作分为:在末尾插入一个字符ch和查询不同子串出现次数不小于K的数量: 思路1:SAM在线求解: 对于每次找到将一个字符x插入到SAM之后,我们知道pre[p]所含有的Tx的后缀字符串数目为step[pre[np]]个,那么只需要每次插入之后更新下这些字符串出现的次数cnt即可: 由于Right(fa)与Right(r)没有交集(…
https://vjudge.net/problem/POJ-2513 题解转载自:優YoU  http://user.qzone.qq.com/289065406/blog/1304742541 题意 给定一些木棒,木棒两端都涂上颜色,求是否能将木棒首尾相接,连成一条直线,要求不同木棒相接的一边必须是相同颜色的. 分析 可以用图论中欧拉路的知识来解这道题,首先可以把木棒两端看成节点,把木棒看成边,这样相同的颜色就是同一个节点 问题便转化为: 给定一个图,是否存在“一笔画”经过涂中每一点,以及经…
Kruskal是有关于最小生成树的算法. 这个算法非常好理解,用一句话来概括就是: 从小到大找不同集合的边. 那么,具体是怎样的呢. 1.先把所有顶点初始化为一个连通分量. 2.从所有边中选择最小的(指权值)边,判断该边是否已经在当前构造的连通分量中,如果是,则放弃这条边,找下一条边. 3.重复第二步,直到所有点都被纳入到当前构造的连通分量中. 所以,这样的算法涉及到了查询当前边是否在集合中和合并一条边到集合中. 于是想到了并查集. 先贴并查集代码,并查集的思想是用一个父亲数组:fa[i]代表节…
Cube Stacking Description Farmer John and Betsy are playing a game with N (1 <= N <= 30,000)identical cubes labeled 1 through N. They start with N stacks, each containing a single cube. Farmer John asks Betsy to perform P (1<= P <= 100,000) op…
Find them, Catch them Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 31102   Accepted: 9583 Description The police office in Tadu City decides to say ends to the chaos, as launch actions to root up the TWO gangs in the city, Gang Dragon…
还是畅通project Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 26860    Accepted Submission(s): 11985 Problem Description 某省调查乡村交通状况,得到的统计表中列出了随意两村庄间的距离.省政府"畅通project"的目标是使全省不论什么两个村庄间都能够实现公路交…
H - Roads not only in Berland Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 25D Description Berland Government decided to improve relations with neighboring countries. First of all, it…
题目链接:http://poj.org/problem?id=2912 题目: 题目大意: n个人进行m轮剪刀石头布游戏(0<n<=500,0<=m<=2000) 接下来m行形如x,y,ch的输入,ch='='表示x,y平局, ch='>'表示x赢y,ch='<'表示x输y, 但是我们不知道x,y的手势是什么; 其中有一个人是裁判,它可以出任意手势,其余人手势相同的分一组,共分为三组,可以存 在空组. 也就是说除了裁判外,其余人每一次出的手 势都相同,问能不能确定裁判是…
Network Time Limit: 1000MS Memory Limit: 30000K Total Submissions: Accepted: Special Judge Description Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can…
http://poj.org/problem?id=2421 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 24132   Accepted: 10368 Description There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can con…
时限: 1000MS   内存限制: 10000K 提交总数: 37001   接受: 17398 描述 热带岛屿拉格里山的首长有个问题.几年前,大量的外援花在了村庄之间的额外道路上.但是丛林不断地超越道路,因此庞大的道路网太昂贵而无法维护.老年人理事会必须选择停止维护一些道路.左上方的地图显示了目前正在使用的所有道路,以及每月维护这些道路的费用.当然,即使路线不像以前那么短,也需要采取某种方式在所有村庄之间保持通行.长老院长想告诉长老委员会每月要花多少钱才能维持连接所有村庄的道路.在上面的地图…
The Head Elder of the tropical island of Lagrishan has a problem. A burst of foreign aid money was spent on extra roads between villages some years ago. But the jungle overtakes roads relentlessly, so the large road network is too expensive to mainta…
题意:给定一个N个节点的树,1<=N<=50000 每个节点都有一个权值,代表商品在这个节点的价格.商人从某个节点a移动到节点b,且只能购买并出售一次商品,问最多可以产生多大的利润. 思路:路径压缩,得到每个点到当前根的信息,然后更新即可. 有可以用倍增做. 很久前抄的代码. #include<cstdio> #define min(a,b) (a<b?a:b) #define max(a,b) (a>b?a:b) #define swap(a,b) (a^=b,b^=…